\(\int \frac {(a+a \sin (e+f x))^m (A+C \sin ^2(e+f x))}{\sqrt {c+d \sin (e+f x)}} \, dx\) [13]

Optimal result

Integrand size = 39, antiderivative size = 364 \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=-\frac {2 C \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {c+d \sin (e+f x)}}{d f (3+2 m)}-\frac {2^{\frac {1}{2}+m} a (2 c (C+2 C m)+d (C-2 C m+A (3+2 m))) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-1+m} \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{d f (3+2 m) \sqrt {c+d \sin (e+f x)}}-\frac {2^{\frac {5}{2}+m} C (d m-c (1+m)) \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {3}{2},\frac {1}{2} (1-\sin (e+f x)),\frac {d (1-\sin (e+f x))}{c+d}\right ) \cos (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^m \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{d f (3+2 m) \sqrt {c+d \sin (e+f x)}} \] Output:

-2*C*cos(f*x+e)*(a+a*sin(f*x+e))^m*(c+d*sin(f*x+e))^(1/2)/d/f/(3+2*m)-2^(1 
/2+m)*a*(2*c*(2*C*m+C)+d*(C-2*C*m+A*(3+2*m)))*AppellF1(1/2,1/2,1/2-m,3/2,d 
*(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(1+sin(f*x+e))^(1/2-m 
)*(a+a*sin(f*x+e))^(-1+m)*((c+d*sin(f*x+e))/(c+d))^(1/2)/d/f/(3+2*m)/(c+d* 
sin(f*x+e))^(1/2)-2^(5/2+m)*C*(d*m-c*(1+m))*AppellF1(1/2,1/2,-1/2-m,3/2,d* 
(1-sin(f*x+e))/(c+d),1/2-1/2*sin(f*x+e))*cos(f*x+e)*(1+sin(f*x+e))^(-1/2-m 
)*(a+a*sin(f*x+e))^m*((c+d*sin(f*x+e))/(c+d))^(1/2)/d/f/(3+2*m)/(c+d*sin(f 
*x+e))^(1/2)
 

Mathematica [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx \] Input:

Integrate[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c + d*Sin[e 
 + f*x]],x]
 

Output:

Integrate[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c + d*Sin[e 
 + f*x]], x]
 

Rubi [A] (warning: unable to verify)

Time = 1.18 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3042, 3525, 27, 3042, 3466, 3042, 3267, 157, 27, 156, 155}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^m \left (A+C \sin (e+f x)^2\right )}{\sqrt {c+d \sin (e+f x)}}dx\)

\(\Big \downarrow \) 3525

\(\displaystyle \frac {2 \int \frac {(\sin (e+f x) a+a)^m (a (A d (2 m+3)+C (d+2 c m))+2 a C (d m-c (m+1)) \sin (e+f x))}{2 \sqrt {c+d \sin (e+f x)}}dx}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m (a (A d (2 m+3)+C (d+2 c m))+2 a C (d m-c (m+1)) \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}dx}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(\sin (e+f x) a+a)^m (a (A d (2 m+3)+C (d+2 c m))+2 a C (d m-c (m+1)) \sin (e+f x))}{\sqrt {c+d \sin (e+f x)}}dx}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 3466

\(\displaystyle \frac {a (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c+d \sin (e+f x)}}dx+2 C (d m-c (m+1)) \int \frac {(\sin (e+f x) a+a)^{m+1}}{\sqrt {c+d \sin (e+f x)}}dx}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \int \frac {(\sin (e+f x) a+a)^m}{\sqrt {c+d \sin (e+f x)}}dx+2 C (d m-c (m+1)) \int \frac {(\sin (e+f x) a+a)^{m+1}}{\sqrt {c+d \sin (e+f x)}}dx}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 3267

\(\displaystyle \frac {\frac {a^3 \cos (e+f x) (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}+\frac {2 a^2 C (d m-c (m+1)) \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {a-a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f \sqrt {a-a \sin (e+f x)} \sqrt {a \sin (e+f x)+a}}}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 157

\(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {\sqrt {2} a^2 C (d m-c (m+1)) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {\sqrt {2} (\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}+\frac {2 a^2 C (d m-c (m+1)) \sqrt {1-\sin (e+f x)} \cos (e+f x) \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a}}}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 156

\(\displaystyle \frac {\frac {a^3 \sqrt {1-\sin (e+f x)} \cos (e+f x) (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \int \frac {(\sin (e+f x) a+a)^{m-\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}+\frac {2 a^2 C (d m-c (m+1)) \sqrt {1-\sin (e+f x)} \cos (e+f x) \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \int \frac {(\sin (e+f x) a+a)^{m+\frac {1}{2}}}{\sqrt {1-\sin (e+f x)} \sqrt {\frac {c}{c-d}+\frac {d \sin (e+f x)}{c-d}}}d\sin (e+f x)}{f (a-a \sin (e+f x)) \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

\(\Big \downarrow \) 155

\(\displaystyle \frac {\frac {\sqrt {2} a^2 \sqrt {1-\sin (e+f x)} \cos (e+f x) (d (A (2 m+3)-2 C m+C)+2 c (2 C m+C)) (a \sin (e+f x)+a)^m \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \operatorname {AppellF1}\left (m+\frac {1}{2},\frac {1}{2},\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) (a-a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}+\frac {2 \sqrt {2} a C (d m-c (m+1)) \sqrt {1-\sin (e+f x)} \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \sqrt {\frac {c+d \sin (e+f x)}{c-d}} \operatorname {AppellF1}\left (m+\frac {3}{2},\frac {1}{2},\frac {1}{2},m+\frac {5}{2},\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+3) (a-a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}}{a d (2 m+3)}-\frac {2 C \cos (e+f x) (a \sin (e+f x)+a)^m \sqrt {c+d \sin (e+f x)}}{d f (2 m+3)}\)

Input:

Int[((a + a*Sin[e + f*x])^m*(A + C*Sin[e + f*x]^2))/Sqrt[c + d*Sin[e + f*x 
]],x]
 

Output:

(-2*C*Cos[e + f*x]*(a + a*Sin[e + f*x])^m*Sqrt[c + d*Sin[e + f*x]])/(d*f*( 
3 + 2*m)) + ((Sqrt[2]*a^2*(2*c*(C + 2*C*m) + d*(C - 2*C*m + A*(3 + 2*m)))* 
AppellF1[1/2 + m, 1/2, 1/2, 3/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e 
 + f*x]))/(c - d))]*Cos[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x 
])^m*Sqrt[(c + d*Sin[e + f*x])/(c - d)])/(f*(1 + 2*m)*(a - a*Sin[e + f*x]) 
*Sqrt[c + d*Sin[e + f*x]]) + (2*Sqrt[2]*a*C*(d*m - c*(1 + m))*AppellF1[3/2 
 + m, 1/2, 1/2, 5/2 + m, (1 + Sin[e + f*x])/2, -((d*(1 + Sin[e + f*x]))/(c 
 - d))]*Cos[e + f*x]*Sqrt[1 - Sin[e + f*x]]*(a + a*Sin[e + f*x])^(1 + m)*S 
qrt[(c + d*Sin[e + f*x])/(c - d)])/(f*(3 + 2*m)*(a - a*Sin[e + f*x])*Sqrt[ 
c + d*Sin[e + f*x]]))/(a*d*(3 + 2*m))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[b/(b*e - a*f)]^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/ 
(b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], x] /; FreeQ[{a, b, c, d, e, f, 
 m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 157

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(c + d*x)^FracPart[n]/(Simplify[b/(b*c - a*d)]^IntPart[n 
]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c 
 - a*d)) + b*d*(x/(b*c - a*d)), x]^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
&  !GtQ[Simplify[b/(b*c - a*d)], 0] &&  !SimplerQ[c + d*x, a + b*x] &&  !Si 
mplerQ[e + f*x, a + b*x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3267

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^2*(Cos[e + f*x]/(f*Sqrt[a + b*Sin[e 
+ f*x]]*Sqrt[a - b*Sin[e + f*x]]))   Subst[Int[(a + b*x)^(m - 1/2)*((c + d* 
x)^n/Sqrt[a - b*x]), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m 
, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && 
 !IntegerQ[m]
 

rule 3466

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A*b - a*B)/b   Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x 
] + Simp[B/b   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ 
[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && NeQ[A*b + a*B, 0]
 

rule 3525

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 
)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2))   Int[(a + b*Sin[e + f* 
x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 
)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
 e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] &&  !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
 

Maple [F]

\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \sin \left (f x +e \right )^{2}\right )}{\sqrt {c +d \sin \left (f x +e \right )}}d x\]

Input:

int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x)
 

Output:

int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x)
 

Fricas [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x, 
algorithm="fricas")
 

Output:

integral(-(C*cos(f*x + e)^2 - A - C)*(a*sin(f*x + e) + a)^m/sqrt(d*sin(f*x 
 + e) + c), x)
 

Sympy [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\sqrt {c + d \sin {\left (e + f x \right )}}}\, dx \] Input:

integrate((a+a*sin(f*x+e))**m*(A+C*sin(f*x+e)**2)/(c+d*sin(f*x+e))**(1/2), 
x)
 

Output:

Integral((a*(sin(e + f*x) + 1))**m*(A + C*sin(e + f*x)**2)/sqrt(c + d*sin( 
e + f*x)), x)
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x, 
algorithm="maxima")
 

Output:

integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/sqrt(d*sin(f*x + e 
) + c), x)
                                                                                    
                                                                                    
 

Giac [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int { \frac {{\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {d \sin \left (f x + e\right ) + c}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x, 
algorithm="giac")
 

Output:

integrate((C*sin(f*x + e)^2 + A)*(a*sin(f*x + e) + a)^m/sqrt(d*sin(f*x + e 
) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c+d\,\sin \left (e+f\,x\right )}} \,d x \] Input:

int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c + d*sin(e + f*x))^( 
1/2),x)
 

Output:

int(((A + C*sin(e + f*x)^2)*(a + a*sin(e + f*x))^m)/(c + d*sin(e + f*x))^( 
1/2), x)
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx=\left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m} \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right ) d +c}d x \right ) c +\left (\int \frac {\sqrt {\sin \left (f x +e \right ) d +c}\, \left (a +a \sin \left (f x +e \right )\right )^{m}}{\sin \left (f x +e \right ) d +c}d x \right ) a \] Input:

int((a+a*sin(f*x+e))^m*(A+C*sin(f*x+e)^2)/(c+d*sin(f*x+e))^(1/2),x)
 

Output:

int((sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a)**m*sin(e + f*x)**2)/(si 
n(e + f*x)*d + c),x)*c + int((sqrt(sin(e + f*x)*d + c)*(sin(e + f*x)*a + a 
)**m)/(sin(e + f*x)*d + c),x)*a